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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2025 · 4 April · Morning Shift · Chemistry · Thermodynamics (C)

One mole of an ideal gas expands isothermally and reversibly from...

One mole of an ideal gas expands isothermally and reversibly from 10dm310 \mathrm{dm}^3 to 20dm320 \mathrm{dm}^3 at 300 K.ΔU300 \mathrm{~K}. \Delta \mathrm{U}, q and work done in the process respectively are :
Given : R=8.3JK−1andmol−1\mathrm{R}=8.3 \mathrm{JK}^{-1} and \mathrm{mol}^{-1}
<br/> In 10=2.3<br>log⁡2=0.30<br>log⁡3=0.48<br/>\begin{aligned}<br />& \text{ In } 10=2.3 \\<br>& \log 2=0.30 \\<br>& \log 3=0.48<br />\end{aligned}

Options

  1. A0,21.84 kJ,−1.26 kJ0,21.84 \mathrm{~kJ},-1.26 \mathrm{~kJ}
  2. B0,−17.18 kJ,1.718 J0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}
  3. C0,21.84 kJ,21,84 kJ0,21.84 \mathrm{~kJ}, 21,84 \mathrm{~kJ}
  4. D0,178 kJ,−1.718 kJ0,178 \mathrm{~kJ},-1.718 \mathrm{~kJ}

Correct answer

D. 0,178 kJ,−1.718 kJ0,178 \mathrm{~kJ},-1.718 \mathrm{~kJ}

Step-by-step solution

(10 L,300 K)→n=1(20 L,300 K)−q=w=−nRTln⁡V2 V1=−8.3×300×ln⁡(2010)=−1.718 kJ⇒q=1.718 kJw=−1.718 kJΔU=0(∵Δ T=0)\begin{aligned} & (10 \mathrm{~L}, 300 \mathrm{~K}) \xrightarrow{\mathrm{n}=1}(20 \mathrm{~L}, 300 \mathrm{~K}) \\ & -\mathrm{q}=\mathrm{w}=-\mathrm{nRT} \ln \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\ & =-8.3 \times 300 \times \ln \left(\frac{20}{10}\right) \\ & =-1.718 \mathrm{~kJ} \\ & \Rightarrow \mathrm{q}=1.718 \mathrm{~kJ} \\ & \mathrm{w}=-1.718 \mathrm{~kJ} \\ & \Delta \mathrm{U}=0(\because \Delta \mathrm{~T}=0)\end{aligned}
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